Consider the reaction $$ 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) $$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K}\) ? c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Step by step solution

01

Calculate ΔGf° for all the reactants and products in the equation

We have the ΔGf° values for POCl3(g) and PCl3(g). Let's write them down: ΔGf°(POCl3) = -502 kJ/mol ΔGf°(PCl3) = -270 kJ/mol We can calculate ΔGf° for O2(g) as it is in its standard state (an element in its most stable form under standard conditions), and which has a ΔGf° value of 0. ΔGf°(O2) = 0 kJ/mol

02

Calculate ΔG° for the reaction

Using the stoichiometric coefficients in the balanced equation and the calculated ΔGf° values, we can calculate the ΔG° for the reaction: ΔG° = (2ΔG°(PCl3) + ΔG°(O2)) - (2ΔG°(POCl3)) Substitute the values: ΔG° = (2 × (-270 kJ/mol) + 0 kJ/mol) - (2 × (-502 kJ/mol)) ΔG° = -540 kJ/mol + 1004 kJ/mol ΔG° = 464 kJ/mol So the change in standard Gibbs free energy (ΔG°) for this reaction is 464 kJ/mol. #b. Is this reaction spontaneous under standard conditions at 298 K?#

03

Determine if the reaction is spontaneous at 298 K

A reaction is spontaneous under standard conditions if the ΔG° for the reaction is negative. In our case, ΔG° = 464 kJ/mol, which is positive. Therefore, the reaction is NOT spontaneous under standard conditions at 298 K. #c. At what temperatures is this reaction spontaneous at standard conditions?#

04

Calculate ΔH° for the reaction

Using the fact that ΔG° = ΔH° - TΔS°, we can solve for ΔH°. We know that ΔS° = 179 J/mol·K, and we already found ΔG° = 464 kJ/mol. Thus, we have: 464 kJ/mol = ΔH° - (298 K) × (179 J/mol·K) Note that we should convert ΔG° to J/mol to match the units of ΔS°: 464 kJ/mol × (1000 J/1 kJ) = 464000 J/mol Now, we solve for ΔH°: ΔH° = 464000 J/mol + (298 K × 179 J/mol·K) ΔH° = 464000 J/mol + 53342 J/mol ΔH° = 517342 J/mol So the change in standard enthalpy (ΔH°) for this reaction is 517342 J/mol.

05

Find the temperature range for spontaneity

Recall the equation ΔG° = ΔH° - TΔS°. A reaction is spontaneous when ΔG° < 0. Therefore, we need to find a temperature for which this inequality holds: ΔH° - TΔS° < 0 We already found that ΔH° = 517342 J/mol and ΔS° =179 J/mol·K. Substitute these values: 517342 J/mol - T × 179 J/mol·K < 0 Now, isolate T to find the temperature range: T > (517342 J/mol)/(179 J/mol·K) T > 2890 K Hence, the reaction is spontaneous at standard conditions for temperatures greater than 2890 K.

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